int(sin(x)*ln(1+sin(x))dx) (I am using int() instead symbol ∫)
apply integration by parts:
Using the product rule, we have that
int(udv)=uv-int(vdu)
let u=ln(1+sin(x)) and v=sin(x)dx
Then du=(ln(sin(x)+1))′dx=(cos(x)/(sin(x)+1))dx
and
v=int(sin(x)dx)=-cos(x)
The integral can be rewritten as
int(ln(sin(x)+1)sin(x)dx)
=(ln(sin(x)+1)*(-cos(x))-int(-cos(x))*(cos(x)/(sin(x)+1))dx)
=(-ln(sin(x)+1)cos(x)-int((-cos^2(x)/(sin(x)+1))dx)
Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:
-ln(sin(x)+1)cos(x)- int((-cos^2(x)/(sin(x)+1))dx)
=-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
Integrate term by term:
-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
=-ln(sin(x)+1)cos(x)-(-int(1dx)+int(sin(x)dx)
Apply the constant rule int(c)dx=cx with c=1:
-ln(sin(x)+1)cos(x)-int(sin(x)dx)+int(1dx)
=-ln(sin(x)+1)cos(x)-int(sin(x)dx)+x
The integral of the sine is int(sin(x)dx)=-cos(x):
x-ln(sin(x)+1)cos(x)-int(sin(x)dx)=x-ln(sin(x)+1)cos(x)-(-cos(x))
Therefore,
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)
Add the constant of integration:
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
Answer:
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
